$u=\dfrac{v_0}{\left(1-\dfrac{v_0^2}{v'^2}\right)^{-1/2}-1}=\dfrac{2.0}{\frac{1}{0.6}-1}=3.0\text{ km/h}$
(ii) Problem Statement
"Two swimmers leave point A on one bank of the river to reach point B lying right across on the other bank. One of them crosses the river along the straight line AB while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point B. What was the velocity $u$ of his walking if both swimmers reached the destination simultaneously? The stream velocity $v_0=2.0$ km/hour and the velocity $v'$ of each swimmer with respect to water equals 2.5 km per hour."
Reference Answer Verification (from the supplied solution image):
The reference finds $t_1=\dfrac{d}{\sqrt{v'^2-v_0^2}}$ for swimmer 1, and $t_2=\dfrac{d}{v'}$, $x=v_0t_2=\dfrac{v_0}{v'}d$ and $t_3=\dfrac{x}{u}$ for swimmer 2. Setting $t_1=t_2+t_3$ gives $u=\dfrac{v_0}{\left(1-v_0^2/v'^2\right)^{-1/2}-1}=3$ km/h. Our work matches.
(iii) The Big Idea in Plain Words
A river carries everything in it downstream. A swimmer moves with the water and also through the water.
Symbols: $d$ is the width of the river. $v_0$ is the speed of the stream. $v'$ is a swimmer's speed relative to the water. $v_1$ is swimmer 1's speed across the river, as seen from the bank. $x$ is how far the stream carries swimmer 2. $u$ is swimmer 2's walking speed on the bank.
Swimmer 1 aims a little upstream. The stream cancels the sideways part, so he goes straight along AB. Swimmer 2 aims straight across. The stream drags him to point C, a distance $x$ from B. He walks back to B. Both arrive together, so the two total times are equal. That one equation gives $u$.
(iv) The Solution Viewed in Each Reference Frame
1. Ground or Lab Frame (Observer on the Bank)
Origin fixed at AWhy: swimmer 1's sideways speed must cancel the stream, so $v'$ is split into two parts.
Along the bank: $v'\sin\alpha=v_0$ (the stream is cancelled). Across: $v_1=\sqrt{v'^2-v_0^2}=\sqrt{2.5^2-2.0^2}=1.5$ km/h. So $t_1=\dfrac{d}{v_1}=\dfrac{d}{\sqrt{v'^2-v_0^2}}$.
Why: swimmer 2 points straight across, so his across speed is the full $v'$.
$t_2=\dfrac{d}{v'}$. The stream carries him $x=v_0t_2=\dfrac{v_0}{v'}d=0.8d$ along the bank to C.
Why: he now walks the distance $x$ back to B at speed $u$.
$t_3=\dfrac{x}{u}=\dfrac{v_0d}{v'u}$.
Why: both reach B at the same time, so the total times are equal.
$$t_1=t_2+t_3\;\Rightarrow\;\frac{d}{\sqrt{v'^2-v_0^2}}=\frac{d}{v'}+\frac{v_0d}{v'u}\;\Rightarrow\;u=\frac{v_0\sqrt{v'^2-v_0^2}}{v'-\sqrt{v'^2-v_0^2}}=3.0\text{ km/h}$$
The width $d$ cancels, so the answer does not depend on it.
2. Water Frame (Medium Frame, Observer Drifting With the Stream)
Origin moves at $v_0$Why: in this frame the water is at rest, so each swimmer moves at exactly $v'$ in the direction he points. The banks, A and B drift upstream at $v_0$.
Swimmer 1's path is slanted, so that he reaches B as B drifts toward him. His speed across is still $\sqrt{v'^2-v_0^2}$, so $t_1=\dfrac{d}{\sqrt{v'^2-v_0^2}}$.
Why: swimmer 2 points straight across, so in this frame he goes straight across at $v'$.
$t_2=\dfrac{d}{v'}$. In that time, B drifts upstream by $v_0t_2=x$. So swimmer 2 ends up a distance $x$ from B, as before.
Why: walking at $u$ on the bank means $u+v_0$ upstream in this frame. B also moves upstream at $v_0$.
The walker closes the gap to B at speed $(u+v_0)-v_0=u$. So $t_3=\dfrac{x}{u}$. Setting $t_1=t_2+t_3$ gives the same $u=3.0$ km/h.
3. Centre-of-Mass (CM) Frame
This frame does not apply. The swimmers move independently, no masses are given, and a centre of mass of the two would add nothing to a question about times and walking speed.
What Changes and What Stays the Same Across Frames
- What changes: the paths and velocities. In the water frame swimmer 2's swim is straight across, and the walker's speed looks like $u+v_0$.
- What stays the same: $d$, the times $t_1,t_2,t_3$, the gap $x$ at the moment of landing, and the walking speed relative to the bank, $u=3.0$ km/h.
(v) The Solution by Each Method
Method 1: Relative-Motion Method (Water Frame)
Fastest- Why: in the water frame, swimmer 2 goes straight across at $v'$, and the stream only moves B away.
$t_2=\dfrac{d}{v'}$ and $x=v_0t_2=0.8d$. - Why: swimmer 1's across speed is the part of $v'$ left after cancelling $v_0$.
$t_1=\dfrac{d}{\sqrt{v'^2-v_0^2}}=\dfrac{d}{1.5}$. - Why: the walking time is whatever time is left.
$t_3=t_1-t_2=0.267d$, so $u=\dfrac{x}{t_3}=\dfrac{0.8d}{0.267d}=3.0$ km/h.
Method 2: Graph Method (Vector Triangles and Timeline)
- Why: for swimmer 1, $\vec v'$, $\vec v_0$ and $\vec v_1$ form a right triangle with $v'$ as the longest side.
$v_1=\sqrt{v'^2-v_0^2}=1.5$ km/h, so $t_1=\dfrac{d}{1.5}$. - Why: for swimmer 2, the triangle with sides $v'$ (across) and $v_0$ (along) is similar to the big triangle $A,\,B,\,C$ in the figure.
So $\dfrac{x}{d}=\dfrac{v_0}{v'}=0.8$. - Why: on the timeline at the bottom of the simulation, the bar of swimmer 1 equals the swim bar plus the walk bar of swimmer 2.
$t_1=t_2+t_3$ with $t_2=\dfrac{d}{2.5}$ and $t_3=\dfrac{0.8d}{u}$. This gives $u=3.0$ km/h.
Method 3: Calculus Method
- Why: the distance across is the integral of the speed across, and it must reach $d$.
Swimmer 1: $\displaystyle\int_0^{t_1}\sqrt{v'^2-v_0^2}\,dt=d\Rightarrow t_1=\dfrac{d}{\sqrt{v'^2-v_0^2}}$. Swimmer 2: $\displaystyle\int_0^{t_2}v'\,dt=d\Rightarrow t_2=\dfrac{d}{v'}$. - Why: the drift is the integral of the stream speed over the swimming time.
$x=\displaystyle\int_0^{t_2}v_0\,dt=v_0t_2$. Walking back: $\displaystyle\int_0^{t_3}u\,dt=x\Rightarrow t_3=\dfrac{x}{u}$. - Why: use $t_1=t_2+t_3$ and solve for $u$.
$u=\dfrac{x}{t_1-t_2}=3.0$ km/h.
Methods 4 and 5: Force and Energy Methods
Force Method: not suited. All speeds are constant, so no net force is involved.
Energy Method: not suited. No masses or work are given.
Fastest: Method 1 (water frame). Most reliable in an exam: Method 3, since writing the three times first and then the one equation $t_1=t_2+t_3$ avoids slips. All give $u=3.0$ km/h.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|---|---|---|
| $d$ | width of the river (not given, cancels out) | km | |
| $v_0$ | stream speed | km/h | |
| $v'$ | swimmer speed relative to the water | km/h | |
| $v_1$ | swimmer 1's speed across, $\sqrt{v'^2-v_0^2}$ | km/h | |
| $t_1,\ t_2,\ t_3$ | swimmer 1's time; swimmer 2's swim time and walk time | min | |
| $x$ | drift of swimmer 2, $v_0d/v'$ | km | |
| $u$ | walking speed relative to the bank (answer) | km/h |
(vii) Common Exam Traps
- Using $v'-v_0$ or $v'$ for swimmer 1's crossing speed. It is $\sqrt{v'^2-v_0^2}$.
- Pointing swimmer 1 straight at B. He must aim upstream.
- Letting the stream change swimmer 2's crossing time. He points straight across, so $t_2=\dfrac{d}{v'}$.
- Taking $u$ as speed relative to the water. It is relative to the bank (the ground).
- Looking for the width $d$. It cancels from the equation.
- Needing $v_0
Verified: the live numbers, both frames and all three methods give $u=3.0$ km/h by default, with all speeds in km/h and times in minutes.